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Choose the convention ${\psi }^{\prime }=g\psi$ for a field in the defining representation and $D_{\mu }={\partial }_{\mu }+A_{\mu }$. Requiring $D_{\mu }^{\prime }{\psi }^{\prime }=g{D}_{\mu }\psi$ gives the Yang-Mills gauge transformationThe gauge field strength is the curvature $[D_{\mu },D_{\nu }]=F_{\mu \nu }$, soFor an adjoint field $Y$, the adjoint covariant derivative is $D_{\mu }Y={\partial }_{\mu }Y+[A_{\mu },Y]$, and direct substitution gives $D_{\mu }^{\prime }{Y}^{\prime }=g(D_{\mu }Y)g^{-1}$. Consequently $D_{\mu }{F}_{\nu \rho }$ transforms by conjugation too.

The Killing-form invariance under a Lie-group adjoint action gives $\kappa (gY{g}^{-1},gZ{g}^{-1})=\kappa (Y,Z)$. Its proof is that the adjoint matrices are conjugated by the same invertible map, leaving their trace product unchanged. With $s=\kappa (F_{\mu \nu },F^{\mu \nu })$, both $s^p$ and $\kappa (D_{\mu }{F}_{\nu \rho },D^{\mu }{F}^{\nu \rho })$ are therefore gauge invariant.

For their field equations, take a compactly supported variation $a_{\mu }=\delta A_{\mu }$. Curvature varies byThe invariant form permits gauge-covariant integration by parts, because ${\partial }_{\mu }\kappa (Y,Z)=\kappa (D_{\mu }Y,Z)+\kappa (Y,D_{\mu }Z)$. Antisymmetry of $F$ givesIntegrating, the power of the Yang-Mills invariant gives the equationFor $p=1$ this recovers the Yang-Mills equations. For larger $p$, keep the displayed product form, including at $s=0$.

For the derivative-squared action, write $B_{\mu \nu \rho }=D_{\mu }{F}_{\nu \rho }$. Varying the derivative as well as the curvature givesIts first variation isIntegrate the first term once, substitute $\delta F_{\nu \rho }=D_{\nu }{a}_{\rho }-D_{\rho }{a}_{\nu }$, and integrate again. It becomes $4\int \kappa (a_{\nu },D_{\mu }{D}^2{F}^{\mu \nu })d^{4}x$, where $D^2=D_{\lambda }{D}^{\lambda }$. Invariance converts the second term into $2\int \kappa (a_{\nu },[F_{\rho \sigma },D^{\nu }{F}^{\rho \sigma }])d^{4}x$. Nondegeneracy of the Killing form and arbitrary variations yield the derivative-squared Yang-Mills action equationThe derivative order matters because covariant derivatives generally do not commute. In an abelian limit the commutator term vanishes and the equation reduces to $\square {\partial }_{\mu }{F}^{\mu \nu }=0$, a useful check on the fourth-order equation for the potential.